Dear Jim,
On May 15, 4:03 am, jimfar jamesfar...@mac.com wrote:
Thanks, I was confusing myself with the definition of the order of an
element with order of the cycle.
Are you really confusing it?
As much as I understood, you only want those elements that have a
single (!) cycle of length 3.
On Fri, May 15, 2009 at 2:36 AM, simon.k...@uni-jena.de wrote:
Dear Jim,
On May 15, 4:03 am, jimfar jamesfar...@mac.com wrote:
Thanks, I was confusing myself with the definition of the order of an
element with order of the cycle.
Are you really confusing it?
As much as I understood,
I must be missing something. Why can't you just check the order of the element?
On Thu, May 14, 2009 at 7:53 PM, jimfar jamesfar...@mac.com wrote:
I can generate a list from any given group, but how would I go about
generating a list of just 3 or 5 cycles?
I can find the order of the element, but I am looking to generate a
list of all of the 3 cycles in something like AlternatingGroup(5)
where the list will not go on for too long.
On May 14, 5:04 pm, David Joyner wdjoy...@gmail.com wrote:
I must be missing something. Why can't you just check the
Why doesn't the obvious 1-liner
[x for x in AlternatingGroup(5) if x.order()==3]
work? Again, am I missing something?
On Thu, May 14, 2009 at 8:57 PM, jimfar jamesfar...@mac.com wrote:
I can find the order of the element, but I am looking to generate a
list of all of the 3 cycles in
Thanks, I was confusing myself with the definition of the order of an
element with order of the cycle.
On May 14, 6:52 pm, David Joyner wdjoy...@gmail.com wrote:
Why doesn't the obvious 1-liner
[x for x in AlternatingGroup(5) if x.order()==3]
work? Again, am I missing something?
On Thu,
